Third Example

Steps 1 and 2

All three terms are already on the left side of the equation, so we
may begin factoring. First, we factor out a greatest common factor
of 3.

3(x4 - 96x2 - 400) = 0

Next, we factor a trinomial.

3(x2 + 4)(x2 - 100) = 0

Finally, we factor the binomial (x2 - 100) as a
difference between two squares.

3(x2 + 4)(x + 10)(x - 10) = 0

Step 3

We proceed by setting each of the four factors equal to zero,
resulting in four new equations.

1. 3 = 0
2. x2 + 4 = 0
3. x + 10 = 0
4. x - 10 = 0

The first equation is invalid, and does not yield a solution. The
second equation cannot be solved using basic methods. (x2
+ 4 = 0 actually has two imaginary number solutions, but we will save Imaginary
Numbers for another lesson!) Equation 3 has a solution of x = -10,
and Equation 4 has a solution of x = 10.

Step 4

We now include all the solutions we found in a single solution to the
original problem: